CODEWITHSUNDEEP
c++variablestypessize
정등혁
정등혁

Reputation: 5

Is there a type for 16 size data?

I heard that in 32bit computer(1990~2000?) had to use a 8 size data like below; (that I heard of... there was no 8 size type, then... ) they used like this;

// 8 size 
typedef union _LARGE_INTEGER {
    struct {
        DWORD LowPart;
        LONG HighPart;
    } DUMMYSTRUCTNAME;
    struct {
        DWORD LowPart;
        LONG HighPart;
    } u;
    LONGLONG QuadPart; // I'm not sure of this part;
} LARGE_INTEGER;

So I want to try to make a 16 size type for my own it is ok to use it like this??

// 16 size 
typedef union INTERGER128 {
    struct {
        LONGLONG LowPart;
        LONGLONG HighPart;
    } u;
} _INTERGER128;

Upvotes: 0

Views: 219

Answers (2)

정등혁
정등혁

Reputation: 5

I used another way to may a 16 size variable

__declspec(align(16)) struct aa {
    long long a;
    long long b;
    char c;
}

this way you can make a 16 size padding in structs

Upvotes: 0

eerorika
eerorika

Reputation: 238301

Is there a type for 16 size data?

There are no standard fundamental types that are guaranteed to be 16 bytes. In some language implementations, there is a 128 bit integer type as a language extension.

typedef union INTERGER128 {
    LONGLONG LowPart;
    LONGLONG HighPart;
} _INTERGER128;

I see no case where your suggested union would be useful. Furthermore, the size of that union would be same as LONGLONG. You haven't shown how that is defined but if it is long long then on most systems that is a 64 bit type; not 128.

You can define a custom 128 bit class like this:

struct INTERGER128 {
    std::uint64_t low;
    std::uint64_t high;
};

P.S. _INTERGER128 and _LARGE_INTEGER identifiers are reserved to the language implementation. You should use another name. Or just don't define a pointless alias.

P.P.S. There is a proposal for standard library based wide integers: p0539

Upvotes: 2

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